how to plot this function
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4 comentarios
darova
el 17 de Mayo de 2019
Do you have any attempts?
HADIMARGO
el 17 de Mayo de 2019
HADIMARGO
el 17 de Mayo de 2019
Walter Roberson
el 18 de Mayo de 2019
(n<-2 & n>2)
is never true. However, since you are multiplying that by 0, if it somehow did match it would contribute 0 anyhow. You can just leave out that term.
Your x axis is wrong. You are effectively using your n as your frequency, including negative n. If you are going to plot negative frequencies then you need to fftshift() . And use more points.
Read the first example of fft() to see how to properly plot.
Respuestas (2)
Sulaymon Eshkabilov
el 18 de Mayo de 2019
n = -5:1:5;
x=1*(n>=-2 & n<=2)+0*(n<-2 & n>2);
L = length(n); Fs = 1;
Nblock = 128; % Block size
Y = fft(x, Nblock)/L;
f = Fs/2*linspace(0,5, Nblock/2+1);
plot(f, 2*abs(Y(1:Nblock/2+1)))
xlabel('frequency, [Hz]'), ylabel('Amplitude')
title('Single-sided Spectrum'), shg
3 comentarios
madhan ravi
el 18 de Mayo de 2019
Editada: madhan ravi
el 18 de Mayo de 2019
x=1*(n>=-2 & n<=2)+0*(n<-2 & n>2); % what is the necessity to multiply it by zero ??
% Hint: Not necessary
HADIMARGO
el 18 de Mayo de 2019
HADIMARGO
el 18 de Mayo de 2019
Sulaymon Eshkabilov
el 18 de Mayo de 2019
Hi,
Here is the full spectrum calculation:
n = -5:1:5;
x=1*(n>=-2 & n<=2);
L = length(n); Fs = 1;
Nblock = 128; % Block size
Y = fft(x, Nblock)/L;
f = Fs/2*linspace(0,5, Nblock);
plot(f, 2*abs(Y(1:Nblock)))
xlabel('frequency, [Hz]'), ylabel('Amplitude')
title('Single-sided Spectrum'), shg
Good luck.
4 comentarios
HADIMARGO
el 18 de Mayo de 2019
Walter Roberson
el 18 de Mayo de 2019
fftshift()
HADIMARGO
el 19 de Mayo de 2019
Walter Roberson
el 19 de Mayo de 2019
t = linspace(0,10);
ft = fft(t);
subplot(1,2,1); plot(abs(ft)); %what you are doing now
subplot(1,2,2); plot(abs(fftshift(ft))); %what you need to do in order to get the 0 in the center
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